You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

The equilibri...

Question

The equilibrium $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ is estabilished in a reaction vessel of $2.5 L$ capacity amount of $N_{2}$ and $O_{2}$ taken at the start were respectively $2$ moles and $4$ moles. Half a mole of nitrate been used up at equilibrium. The molar concentration of nitric oxide is

0.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.4$
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ $V o l = 2.5 L$
At $2$ $4$ $0$
$t = 0$ $2 - x$ $4 - x$ $2 x$
$x = 1 / 2$ given
$3 / 2$ $7 / 2$
concentration $= \frac{m o l e s}{V i n l i t} = \frac{1}{2.5} \Rightarrow 0.4$

Suggest Corrections

Similar questions

Q. The equilibrium

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

established in a reaction vessel of

2.5

L capacity. The amounts of

N_{2}

and

O_{2}

taken at the start were respectively

2

moles and

4

moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is :

Q. The equilibrium

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

is established in a reaction vessel of

2.5 L

capacity. The amounts of

N_{2} and O_{2}

taken at the start were

4 and 6 moles

, respectively, and half a mole of nitrogen was used up at equilibrium. What is the molar concentration of nitric oxide at equilibrium?

Q. The equilibrium established in a reaction vessel of

2.5

L capacity. The amounts of

N_{2}

and

O_{2}

taken at the start were respectively

2

moles and

4

moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is:

N_{2} + O_{2} ⇌ 2 N O

Q. In a reaction vessel of 2 litre capacity 3 moles of

N_{2}

reacts with 2 moles of

O_{2}

to produce 1 mole of NO according to following reaction-

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

. What is the molar concentration of

N_{2}

at equilibrium?

Q. The equilibrium

N_{2} + O_{2} ⇌ 2 N O

is established in a vessel with a capacity of 2.5 L. The initial amount of

N_{2}

is 2 mol and that of

O_{2}

is 4 mol. Half a mole of

N_{2}

is used up at the equilibrium. The molar concentration of NO is

Join BYJU'S Learning Program

The equilibrium N2(g)+O2(g)⇌2NO(g) is estabilished in a reaction vessel of 2.5 L capacity amount of N2 and O2 taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrate been used up at equilibrium. The molar concentration of nitric oxide is

The correct option is B 0.4 N2(g)+O2(g)⇌2NO(g) Vol=2.5LAt 2 4 0t=0 2−x 4−x 2xx=1/2 given 3/2 7/2concentration =molesVinlit=12.5⇒0.4

The equilibrium $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ is estabilished in a reaction vessel of $2.5 L$ capacity amount of $N_{2}$ and $O_{2}$ taken at the start were respectively $2$ moles and $4$ moles. Half a mole of nitrate been used up at equilibrium. The molar concentration of nitric oxide is

The correct option is B $0.4$
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ $V o l = 2.5 L$
At $2$ $4$ $0$
$t = 0$ $2 - x$ $4 - x$ $2 x$
$x = 1 / 2$ given
$3 / 2$ $7 / 2$
concentration $= \frac{m o l e s}{V i n l i t} = \frac{1}{2.5} \Rightarrow 0.4$